The probability that a marksman will hit a target is given as 15. Then the probability that he hits at least once in 10 shots is
A
1−(45)10
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B
1510
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C
1−(1510)
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D
(45)10
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Solution
The correct option is A1−(45)10 Let H be the event of hitting the shot, then P(¯¯¯¯¯H)=1−P(H)=1−15=45
Now let E be the event of hitting at least once in 10 shots
Then P(E)=1−P (no hit in all 10 shots) ⇒P(E)=1−(45)10